prove ((a+b)/2)^n≦(a^m+b^m)/2

I think your proposition should be: ((a+b)/2)^n≦(a^n+b^n)/2
Without loss of generality, assume a≦b
(This is standard A.M. >= G.M. question, but I do not use this rule)

For n=1, (a+b)/2 ≦(a+b)/2 obviously holds

Given for all 1<=k< p: ((a+b)/2)^k≦(a^k+b^k)/2 is true
(a^(p+1)+b^(p+1))/2
= (a^p+b^p)/2) (a+b) - (a^p b + b^p a)/2
= ((a^p+b^p)/2) ((a+b)/2) + ((a^p+b^p)/2) ((a+b)/2) - (a^p b + b^p a)
= ((a^p+b^p)/2) ((a+b)/2) + ((a^p - b^p)/2)((a-b)/2)
>= ((a^p+b^p)/2) ((a+b)/2)
>= ((a+b)/2)^p ((a+b)/2)

The position is proved by induction